Муодилаи тригонометриро ҳал кунед:
\(cos^{-2}2t - sin^{-2}2t = \frac{8}{3}\).
Ҳал.
1)\(cos^{-2}2t = \frac{1}{cos^{2}2t}=1+tg^{2}2t\);
2)\(sin^{-2}2t = \frac{1}{sin^{2}2t}=1+ctg^{2}2t\).
Аз ин ду баробарӣ мебарояд, ки:
\(cos^{-2}2t - sin^{-2}2t=1+tg^{2}2t-\)
\(-(1+ctg^{2}2t)=1+tg^{2}2t-1-ctg^{2}2t=\)
\(=tg^{2}2t-сtg^{2}2t\).
Яъне, \(tg^{2}2t-сtg^{2}2t=\frac{8}{3}\).
3)\(ctg^{2}2t=\frac{1}{tg^{2}2t}\).
\(tg^{2}2t-\frac{1}{tg^{2}2t}=\frac{8}{3}\)
\(\frac{tg^{4}2t}{tg^{2}2t}-\frac{1}{tg^{2}2t}=\frac{8}{3}\)
\(\frac{tg^{4}2t-1}{tg^{2}2t}=\frac{8}{3}\)
\(3\cdot(tg^{4}2t-1)=8tg^{2}2t\)
\(3tg^{4}2t-8tg^{2}2t-3=0\)
Бигзор \(x = tg^{2}2t\). Пас,
\(3x^{2}-8x-3=0\)
\(D=b^2-4\cdot a\cdot c=(-8)^2-4\cdot 3\cdot(-3)=\)
\(\quad=64+36=100=10^2\), \(D>0\)
\(x_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-(-8)\pm \sqrt{10^2}}{2\cdot3}=\frac{8\pm 10}{6}\)
\(x_{1}=\frac{8-10}{6}=\frac{-2}{6}=\frac{-1}{3}=-\frac{1}{3}\);
\(x_{2}=\frac{8+10}{6}=\frac{18}{6}=3\);
Якум ҳолатро дида мебароем:
\(tg^{2}2t=-\frac{1}{3}\)
\(tg 2t=\sqrt{-\frac{1}{3}}\), лекин ин ғайриимкон аст.
Ҳолати дуюмро дида мебароем:
\(tg^{2}2t=3\)
\(tg 2t=\pm \sqrt{3}\)
\(2t=arctg(\pm\sqrt{3})+\pi n\)
\(2t=\pm\frac{\pi}{3}+\pi n\)
\(t=\pm\frac{\pi}{6}+\frac{\pi n}{2}\)
\(t=\frac{\pi}{6}\cdot(3n\pm 1)\).
Ҷавоб: \(t=\frac{\pi}{6}\cdot(3n\pm 1)\), \(n\in \mathbb{Z}\).